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Binomial expansion

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the binary decimal expansion of the fraction 1/3. Identify the repeating decimals of digits.


    3. The attempt at a solution
    I have that 1/3=0.0101111... and so the repeating digit is 1.

    Is this right? It's the first time I've been exposed to binary expansion.
     
  2. jcsd
  3. Nov 24, 2008 #2
    No. In binary, 0.010111111... = 0.011 = 11/1000 = 3/8dec. What method are you using to find the binary expansion?
     
  4. Nov 24, 2008 #3
    1/3=x(1/2) + x(1/2)^2 + x(1/2)^3...

    where each x is either 0 or 1. The first x cannot be 1 because 1/2 > 1/3
    I used the same reasoning for the other x's.

    How do you do it correctly?
     
  5. Nov 24, 2008 #4
  6. Nov 24, 2008 #5
    What am I supposed to divide by though? And what does this mean from the website you linked:
    "33 / 3 (i.e. 100001 / 11)"
     
  7. Nov 24, 2008 #6
    You divide 1bin = 1dec by 11bin = 3dec.

    "(i.e. 100001 / 11)" is simply stating that the binary equivalent of 33dec is 100001bin, and that 3dec = 11bin, so 33/3dec = 100001/11bin
     
  8. Nov 25, 2008 #7

    HallsofIvy

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    I would do this by saying: 1/3 is less than 1/2 but larger than 1/4: 1/3- 1/4= 1/12 so 1/3= 1/4+ 1/12. 1/12 is less than 1/8 but larger than 1/16: 1/12- 1/16= 1/48 so 1/3= 1/4+ 1/16+ 1/48. 1/48 is less than 1/32 but larger than 1/64: 1/48- 1/64= 1/192 so 1/3= 1/4+ 1/16+ 1/64+ 1/192. You should be able to see the pattern now.
     
  9. Nov 25, 2008 #8
    And once you see the pattern, you could always check its correctness by summing the geometric series. I really would try to get a handle on doing arithmetic in binary though, since I think it provides more insight than just looking at the problem in longhand. After all, if the problem was "find the denary decimal expansion of 1/3", would you really set it up by saying

    [tex]\frac{1}{3} = \frac{d_1}{10} + \frac{d_2}{10^2} + \frac{d_3}{10^3} ...[/tex]

    and trying to find the [tex]d_i[/tex], or would you take advantage of the compact notation provided by the division algorithm?
     
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